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7.Gravitation
normal
The orbit of geostationary satellite is circular, the time period of satellite depends on $(i)$ mass of the satellite $(ii)$ mass of the earth $(iii)$ radius of the orbit $(iv)$ height of the satellite from the surface of the earth
A
$(i)$ only
B
$(i)$ and $(ii)$
C
$(i), (ii)$ and $(iii)$
D
$(ii), (iii)$ and $(iv)$
Solution
$\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}=\mathrm{m} \omega^{2}(\mathrm{R}+\mathrm{h})$
$\omega=\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{3}}}=\frac{2 \pi}{\mathrm{T}}$
$T$ depends upon mass of earth's, radius of orbit.
Option is $( 4)$
Standard 11
Physics
